A force $\to F = (3 x y - 5 z) ˆ j + 4 z ˆ k$ $vecF = (3xy -5z)hatj + 4zhatk$ is applied on a particle. The work done by the force when the particle moves from point $(0, 0, 0)$ $(0,0,0)$ to point $(2, 4, 0)$ $(2,4,0)$ along the path $y = x^{2}$ $y = x^2$ is?

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A force $\to F = (3 x y - 5 z) ˆ j + 4 z ˆ k$ $vecF = (3xy -5z)hatj + 4zhatk$ is applied on a particle. The work done by the force when the particle moves from point $(0, 0, 0)$ $(0,0,0)$ to point $(2, 4, 0)$ $(2,4,0)$ along the path $y = x^{2}$ $y = x^2$ is?

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Answer 1 · 2018-02-17T17:06:18

$W = 48 [J]$ $W=48 [J]$

Explanation:

Use the definition of work,

$W = \int \to f \cdot d \to r$ $W = intvecf*dvecr$

Instead using unit vector notations, let's use coordinate to represent the vectors, i.e., $\to f = (0, 3 x y - 5 z, 4 z); and d \to r = (d x, d y, d z$ $vec f = (0, 3xy -5z, 4z); and dvecr = (dx, dy, dz$ )

$W = \int_{(0,0,0)}^{(2,4,0)} (0, 3 x y - 5 z, 4 z) \cdot (d x, d y, d z$ $W = int_("(0,0,0)" )^("(2,4,0)")(0, 3xy -5z, 4z)*(dx, dy, dz$ )#

$W = \int_{(0,0,0)}^{(2,4,0)} (3 x y - 5 z) d y + 4 z d z$ $W = int_("(0,0,0)" )^("(2,4,0)") (3xy -5z)dy +4zdz$
$W = \int_{0}^{4} (3 x y - 5 z) d y + \int_{0}^{0} 4 z d z$ $W= int_0^4 (3xy -5z)dy + int_0^0 4zdz$
$because$ no displacement in z, z-component force contribute no work, hence

$W= [3/2xy^2-5zy]_0^4 + 0$

$W= 24x-20z$
The above is the work done alone the path $y=x^2$ on a z-plane.

At point (2,4, 0),

$W= 24x-20z= 24(2)-20(0)=48$

The units are not specified, hence the work done has not unit, otherwise it is assumed as $W=48J$

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A force $\to F = (3 x y - 5 z) ˆ j + 4 z ˆ k$ $vecF = (3xy -5z)hatj + 4zhatk$ is applied on a particle. The work done by the force when the particle moves from point $(0, 0, 0)$ $(0,0,0)$ to point $(2, 4, 0)$ $(2,4,0)$ along the path $y = x^{2}$ $y = x^2$ is?

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A force vecF = (3xy -5z)hatj + 4zhatk→F=(3xy−5z)ˆj+4zˆkvecF = (3xy -5z)hatj + 4zhatk is applied on a particle. The work done by the force when the particle moves from point (0,0,0)(0,0,0)(0,0,0) to point (2,4,0)(2,4,0)(2,4,0) along the path y = x^2y=x2y = x^2 is?

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A force $\to F = (3 x y - 5 z) ˆ j + 4 z ˆ k$ $vecF = (3xy -5z)hatj + 4zhatk$ is applied on a particle. The work done by the force when the particle moves from point $(0, 0, 0)$ $(0,0,0)$ to point $(2, 4, 0)$ $(2,4,0)$ along the path $y = x^{2}$ $y = x^2$ is?